Question

A coil of current-carrying Nichrome wire is immersed in a liquid contained in a calorimeter. When the potential difference across the coil is 12 V and the current through the coil is 5.2 A, the liquid boils at a steady rate, evaporating at the rate of 21 mg/s. Calculate the heat of vaporization of the liquid (in J/kg).

Answer 1

Potential difference, V=12 v

Current, I=5.2 A

Power=energy /time..... Say time t=1 sec

Power=VI =12*5.2 =62.4 joule/sec.

This is in the form of heat delivered to vaporize the liquid.

As liquid is evaporating at the rate of 21 mg/s by the supplied heat, therefore heat required to vaporize 21 mg =62.4 joule

Heat required to vaporize 1 mg=62.4/21 =2.97 joule.

So heat required to evaporate 1 kg i.e 10^6 mg is

10^6 *2.97 =2.97*10^6 joule.

Thus heat of vaporization is 2.97*10^6 joules /kg.